@PVAL-inactive Wait, first can I clarify something. Is $\{ (x, y) ; -1\leq y\leq 0 \}$ open in $H^2$? So that plus the map $f(a,b) = (\sin a, \cos a, b)$ is enough for a chart?
that being said, I am working on a paper, and fear that I might have to read up on algebraic geometry a bit to figure out how you folk define the term "dimension"
@Adeek I'm too old to learn new tricks
homological algebra will always and forever remain dark voodoo magic to me
can some one help i am trying to learn polynomial long division but i don't understand the calculations going on in these workings: i.imgur.com/wh8KJHz.png why isn't it -x-3 after applying x to cancel x^2
im also not sure i understand where the -9 came from
Like in the example for matrix differentiation: $0 &= \frac{d}{dH_{kj}}[\sum_{ij} (-V_{ij} \sum_{k} \pi_{ijk} log \frac{W_{ik}H_{kj}}{\pi_{ijk}} + \sum_{k}W_{ik}H_{kj})$
after this, it said "for fixed k,j" and then moved on to do differentiation
I want to know how do I know I have to do "fixed k, j"
its just a really nasty thing to calculate involving matrix's and lots of derivatives, they are nicknames christaweful symbols for being so terrible to compute
my prof thought it was reasonable too when he put in on our hw assingment
until he put it into his computer and asked it to calculate it a couple hours before it was due and the computer was still running it when he got to class
i was the only person in the class who actually finished the question
I have come to the conclusion recently that I need to learn more differential geometry :\
the "distance zeta function", an object that is of interest to we fractal geometry-type people, supposedly picks up on thing that are of interest to diff geo-type people, such as some strange thing called "curvature"
and which is off from the geometric zeta function by a constant
@EricSilva Roughly speaking, if $A$ is a bounded subset of $\mathbb{R}^n$, then the distance zeta function associated to $A$ is defined by the integral $$ \zeta_A(s) := \int_{A_{\delta}} d(x,A)^{s-n}\,\mathrm{d}x $$
it turns out that if the upper Minkowski dimension of $A$ is $D$ and the lower Minkowski content of $A$ is positive, then $\zeta_A(s)$ has a singularity at $D$
otherwise, it converges on the right-half plane $\{ \Re(s) > D \}$
and if $\zeta_A$ can be extended to a function that is meromorphic on a strictly larger domain, we refer to the poles (and maybe other singularities?) as the complex dimensions of $A$
@Annelise this is something that you want to think about geometrically. You draw a disk around the origin and take some point outside of it. The closest you'll get is by drawing a line from the point to the origin and finding where it intersects the disk
Now, to formally prove that you can probably make some kind of convexity/symmetry argument?
Like, assume you have some $y\in S$ such that $y \ne tx$
Reflect that point across the line $tx$, draw a line segment there, intersect that line segment with the line $tx$, and the point of intersection should be closer to $x$ than $y$
Sanity-check: If $f$ is a nondecreasing homeomorphism of $[0, 1]$ such that $0$ and $1$ are the only fixed points of $f$, then $0$ is repelling and $1$ is attracting, yeah?
Take a point $x \in [0, 1]$ and look at the iterates $f^k(x)$. That's nondecreasing and bounded by $1$, so converges somewhere
But that point of convergence is a fixed point of $f$, so it has to be $1$
Forward iterates of $f$ converges to $1$ and similarly negative iterates of $f$ converge to $0$
Ok, I'm happy
I guess this means if $f$ is a (orientation preserving) homeomorphism of the circle so that $O_z$ is the one and only periodic orbit of $f$ for some $z \in S^1$, then for any $z' \in O_z$, given a small arc $A = [-\epsilon + z', \epsilon + z']$ around $z'$, $f$ is repelling on the negative side and attracting on the positive side
aka $O_z$ is a semistable orbit
(Because if $z$ is $q$-periodic, it's a fixed point of the homeomorphism $f^q$ of that circle. Cut along $z$ to get a (oriented) homeomorphism of the interval fixing the endpoints: invoke the previous thing)
The semistability is pictorially intriguing, however
In the foliation on the torus given by suspending $f$, the leaf $L$ corresponding to $z$ is semistable: every leaf on one side of $L$ accumulates to $L$ and every leaf on the other side of $L$ repels from $L$
It does prove for every x in [0, 1] either the forward orbit of x converges to 1 and the backward orbit of x converges to 0 or vice versa. I suppose it is not enough to ensure that the roles are the same for every x
hi, suppose $f$ is analytic function s.t $f'$ has zero of order $k$ at $z_0$, i need to show that $f$ can be written as $f(z) = f(z_0) + [g(z)] \ ^ {k+1}$ , someone can help? i managed to show that if a function $h$ has a zero of order $k$ then $h(z) = [l(z)] \ ^ k$, not sure how to continue..
If I have a set in which every point is a limit point can this set be a closed interval $[a,b]$?
It is not specified if the limit points are "right" limit points or "left" limit points. Therefore I assume any interval of the form $[a,b] \in \mathbb R$ is such a set. Is this correct?
so a right limit can be defined on a "right" limit point. Same for left limit. And a "normal" limit point (i.e. right and left limit point) can be used for a "normal" limit.
Anyway back to my question. A set in which every point is a limit point can be a closed interval?
a right limit for a function is something like $\lim_{x \to a^{+}} f(x)$, but a limit point for a set is just a point for which any neighborhood contains other points of your set, there's no meaningful notion of left vs right in the latter but there is in the former
